# Power series which are $p$-adic modular forms for all $p$

Suppose that, for some integer $k$, a series $f(q) \in \mathbb Q \otimes \mathbb Z[[q]]$ has the property that for every prime $p$, $f(q)$ is the $q$-expansion of a $p$-adic modular form of weight $k$ and level $1$. Write $\mathcal M_k$ for the $\mathbb Q$-vector space of such power series, and $\mathcal M$ for the graded sum of the $\mathcal M_k$'s, which has the structure of a graded $\mathbb Q$-algebra.

Every classical modular form of weight $k$ defined over $\mathbb Q$ yields an $f \in \mathcal M_k$, but it turns out that there exist nontrivial examples of "everywhere $p$-adic" modular forms as above, which are not classical modular forms. Serre's operator $\theta = q \frac{d}{dq}$ preserves the space of $p$-adic modular forms for every $p$ (more precisely it takes $p$-adic modular forms of weight $k$ to $p$-adic modular forms of weight $k+2$), and since $\theta$ acts on $q$-expansions in a manner independent from $p$, it preserves $\mathcal M$, even though $\theta f$ is in general not a classical modular form if $f$ is.

More generally, any nearly holomorphic modular form defined over $\mathbb Q$ determines an element of $\mathcal M$, via the map which takes a nearly holomorphic modular form $f_0(\tau) + f_1(\tau)/y + \dots + f_r(\tau)/y^r$ to its "constant term" $f_0$. The function $f_0$ is called a quasimodular form in the terminology of Zagier and Kaneko. Urban shows that the $q$-expansion of $f_0$ is a $p$-adic modular form for all $p$, but $f_0$ is not a classical modular form in general. For instance, the quasimodular Eisenstein series $E_2(q)$ is a $p$-adic modular form of weight $2$ for every $p$, even though the space of classical modular forms of weight $2$ is zero.

I am curious to know whether there are any other examples. Does there exist an $f(q) \in \mathcal M_k$ which is not a quasimodular form? It seems quite likely to me that there is not. If this were true, this would constitute a sort of local-to-global principle. But I have no idea how to attack this problem; I don't even see any reason why $\mathcal M_k$ should be finite-dimensional!

• Same level or varying level? – Will Sawin Feb 10 '15 at 2:00
• @Will, I am happy to have everything in level $1$. – Bruno Joyal Feb 10 '15 at 2:09